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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 10

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 10

Answers (1)

Answer: i=\frac{E}{R}\left[1-e^{\left(-\frac{K}{L}\right) t}\right]

Given:  L \frac{d i}{d t}+R i=E

To find: To prove that i=\frac{E}{R}\left[1-e^{\left(-\frac{R}{L}\right) t}\right]

Hint: Integrate the linear differential equation to find integral factor.

Solution: L \frac{d i}{d t}+R i=E

 Dividing by L

        =\frac{d t}{d t}+\frac{R}{L} i=\frac{E}{L}

This is a linear differential equation and comparing it with

        =\frac{d y}{d x}+P y=Q

Such that,

        =P=\frac{R}{L^{\prime}}, Q=\frac{E}{L}

We know that

        \begin{aligned} &\text { I.F. }=e^{\int p d t} \\\\ &=e^{\int \frac{R}{L} d t} \\\\ &=e^{\left(\frac{R}{L}\right) t} \end{aligned}

So, the solution is given by

        \begin{aligned} &i(I f)=\int Q(I f) d t+C \\\\ &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\int \frac{E}{L}\left[e^{\left(\frac{R}{L}\right) t}\right] d t+C \\\\ &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\frac{E}{L} \times \frac{L}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C \end{aligned}

        \begin{aligned} &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\frac{E}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C \\\\ &=>i=\frac{\frac{E}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C}{\left[e^{\left.\left(\frac{R}{L}\right) t\right]}\right.} \end{aligned}

        \begin{aligned} &\text { => } i=\frac{E}{R}+\frac{C}{\left[e^{\left(\frac{R}{L}\right) t}\right]} \\\\ &=>i=\frac{E}{R}+C\left[e^{-\left(\frac{R}{L}\right) t}\right] \ldots(i) \end{aligned}

Initially no current passes through the circuit so,i=0, t=0

        \begin{aligned} &=>0=\frac{E}{R}+C e^{0} \\\\ &=>0=\frac{E}{R}+C \\\\ &=>C=-\frac{E}{R} \end{aligned}

Substituting values of C in equation (i)

        \begin{aligned} &=>i=\frac{E}{R}-\frac{E}{R}\left[e^{-\left(\frac{R}{L}\right) t}\right] \\\\ &=>i=\frac{E}{R}\left[1-e^{-\left(\frac{R}{L}\right) t}\right] \end{aligned}

 

Posted by

infoexpert26

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